🟨 Construct Binary Tree from Preorder and Inorder Traversal (#105)

🔗 Problem Link

• LeetCode - Construct Binary Tree from Preorder and Inorder Traversal
• NeetCode - Construct Binary Tree from Preorder and Inorder Traversal

📋 Problem Statement

Given two integer arrays preorder and inorder where:

• preorder is the preorder traversal of a binary tree
• inorder is the inorder traversal of the same tree

Construct and return the binary tree.

💡 Examples

Example 1

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2

Input: preorder = [-1], inorder = [-1]
Output: [-1]

🔑 Key Insights & Approach

Core Observation:

• Preorder: root is first element
• Inorder: root splits left and right subtrees
• Use this to recursively build tree

Why Recursive with HashMap?

• O(n) time with hashmap for inorder indices
• O(n) space for hashmap + recursion stack
• Clean recursive structure

Approaches:

1. Recursive with hashmap: O(n) time, O(n) space - optimal
2. Recursive without hashmap: O(n²) time

Pattern: "Tree Construction from Traversals" using divide and conquer.

🐍 Solution: Python

Approach: Recursive with HashMap

Time Complexity: O(n) | Space Complexity: O(n)

from typing import List, Optional

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        # Create hashmap for O(1) inorder lookups
        inorder_map = {val: idx for idx, val in enumerate(inorder)}
        self.pre_idx = 0

        def build(left, right):
            if left > right:
                return None

            # Root is current preorder element
            root_val = preorder[self.pre_idx]
            root = TreeNode(root_val)
            self.pre_idx += 1

            # Find root in inorder to split left/right
            mid = inorder_map[root_val]

            # Build left and right subtrees
            root.left = build(left, mid - 1)
            root.right = build(mid + 1, right)

            return root

        return build(0, len(inorder) - 1)

🔵 Solution: Golang

Approach: Recursive with Map

Time Complexity: O(n) | Space Complexity: O(n)

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func buildTree(preorder []int, inorder []int) *TreeNode {
    inorderMap := make(map[int]int)
    for i, val := range inorder {
        inorderMap[val] = i
    }

    preIdx := 0

    var build func(left, right int) *TreeNode
    build = func(left, right int) *TreeNode {
        if left > right {
            return nil
        }

        rootVal := preorder[preIdx]
        root := &TreeNode{Val: rootVal}
        preIdx++

        mid := inorderMap[rootVal]

        root.Left = build(left, mid-1)
        root.Right = build(mid+1, right)

        return root
    }

    return build(0, len(inorder)-1)
}

📚 References

• NeetCode - Construct Binary Tree from Preorder and Inorder Traversal